# 1. For the spiral r=e^Θ and Θ=t: (a) compute the tangential & normal components of the velocity when t=0. (b) compute the radial & angular components of the acceleration when t=0. 2. A...

1. For the spiral r=e^Θ and Θ=t:

(a) compute the tangential & normal components of the velocity when t=0.

(b) compute the radial & angular components of the acceleration when t=0.

2. A particle moves along the ellipse r=30/(4+cosΘ) in an inverse square gravity field centered at the origin, and it is known that dΘ/dt = 1 when Θ=0.

(a) compute the particle's fastest & slowest speeds in its orbit.

(b) what is the period of the orbit?

*print*Print*list*Cite

### 1 Answer

The figure with the spiral is below.

`r = r^Theta` with `Theta=t`

The parametric equations of the spiral components `x, y` are

`x(t) =r*cos(alpha) =e^t*cos(omega*t)`

`y(t) =r*sin(alpha) =e^t*sin(omega*t)`

with the initial condition: `t=0 rArr Theta=0 => omega*t =0`

and `omega=constant`

In rectangular coordinates written as a vector we have

`r =x*hati +y*hatj =[e^t*cos(omega*t)]*hati +[e^t*sin(omega*t)]*hatj `

a)

The speed of the particle is by definition

`v =(dr)/(dt)`

which gives by components

`(dx)/dt =e^t*cos(omega*t) -omega*e^t*sin(omega*t) =x-omega*y`

`(dy)/dt =e^t*sin(omega*t) +omega*e^t*cos(omega*t) =y +omega*x`

Therefore the speed written as a vector is

`v = (x-omega*y)*hati +(y+omega*x)*hatj`

Since at `t=0` the radius `r` is horizontal, the normal speed is directed in the `hati` (x axis) vector direction, and the tangential speed is directed in the `hatj` (y axis) vector direction

`v_t(t=0) =y(0)+omega*x(0) =omega*e^0 =omega`

`v_n(t=0) =x(0)-omega*y(0) =e^0 =1`

b)

The acceleration is

`a (=(d^2r)/dt^2)=(dv)/dt`

On its components:

`(dv_x)/dt =d/dt(x-omega*y) =(x-omega*y)-omega(y+omega*x) =x-2*omega*y -omega^2*x`

`(dv_y)/dt^2 =d/dt(y+omega*x) =(y+omega*x) +omega(x-omega*y) =y+2*omega*x-omega^2*y`

Therefore the acceleration written as a vector is

`a =(x-2omega*y-omega^2*x)*hati +(y+2omega*x-omega^2*y)*hatj`

At `t=0` the normal component of `a` is in the `hati` (x axis) direction and the tangential component of `a` is in the `hatj` (y axis) direction.

`a_t(0) =y(0)+2*omega*x(0) -omega^2*y(0) =2*omega*e^0 =2*omega`

`a_n(0) =x(0)-2*omega*y(0)-omega^2*x(0) =1-omega^2`

**Answer: the tangential and normal components of the speed, respectively acceleration at `t=0` are `omega` and 1, respectively `2*omega` and `1-omega^2` **